Question

fie familia de parabole fm:R->R
$fm(x) = mx {}^{2} + 2(m - 1)x + m - 1$
;m face parte din R*
a) sa se arate ca varfurile parabolelor satisfac relatia y=-x
b) sa se arate ca parabolele trec printr un punct fix​

Answer 1

$f_m(x) = mx^2+2(m-1)x+m-1\\ \\ a)\quad V\Big(-\dfrac{b}{2a},-\dfrac{\Delta}{4a}\Big)\in (y = -x) \\ \\ \Rightarrow -\dfrac{\Delta}{4a} = -\Big(-\dfrac{b}{2a}\Big) \Rightarrow -\dfrac{4(m-1)^2-4m(m-1)}{4m} = \dfrac{2(m-1)}{2m} \Rightarrow \\ \\ \Rightarrow -\dfrac{4(m-1)\Big((m-1)-m\Big)}{4m} = \dfrac{4(m-1)\cdot 1}{4m} = \dfrac{m-1}{m} \Rightarrow\\ \\\Rightarrrow \dfrac{m-1}{m}= \dfrac{m-1}{m}\quad (A)$

$b)\quad f_m(x) = mx^2+2(m-1)x+m-1\\ \\ f_m(a) = b,\quad f_m(a)\rightarrow \text{nu depinde de m}\\ \\ ma^2+2(m-1)a+m-1 = b \\ ma^2+2ma-2a+m-1 = b\\ m(a^2+2a+1)-2a-1 = b \\ m(a+1)^2-2a-1 = b \\ \\ \Rightarrow (a+1)^2 = 0 \Rightarrow a+1 = 0 \Rightarrow a = -1 \\ \\ \Rightarrow -2\cdot (-1)-1 = b \Rightarrow b = 1 \\ \\ \Rightarrow \text{Toate parabolele trec prin punctul }(-1,1)$